Find $\dfrac{d}{dx}\left(e^{^{\frac{1}{x}}}\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{e^{^{\frac{1}{x}}}}{x^2}$ (Choice B) B $-e^{^{\frac{1}{x}}}$ (Choice C) C $\dfrac{e^{^{\frac{1}{x}}}}{x}$ (Choice D) D $\dfrac{1}{x}\cdot e^{^{\frac{1}{x}-1}}$
Explanation: $e^{^{\frac{1}{x}}}$ is a composition of two, more basic, functions: $\dfrac{1}{x}$ and $e^x$. In other words, suppose $u(x)=\dfrac{1}{x}$ and $v(x)=e^x$, then $e^{^{\frac{1}{x}}}=v\Bigl(u(x)\Bigr)$, or $(v\circ u)(x)$. Therefore, $\dfrac{d}{dx}\left(e^{^{\frac{1}{x}}}\right)$ can be found using the chain rule : $\begin{aligned} \dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&=\dfrac{dv}{du}\cdot\dfrac{du}{dx} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x) \end{aligned}$ Finding $v'\Bigl(u(x)\Bigr)$ $v(x)=e^x$, and therefore $v'(x)=e^x$. Now we plug $u(x)=\dfrac1x$ into $v'$ : $\begin{aligned} v'\Bigl(u(x)\Bigr)&=v'\left(\dfrac{1}{x}\right) \\\\ &={e^{^{\frac{1}{x}}}} \end{aligned}$ Finding $u'(x)$ $u(x)=\dfrac{1}{x}$, and therefore $u'(x)={-\dfrac{1}{x^2}}$. Putting things together $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(e^{^{\frac{1}{x}}}\right) \\\\ &=\dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&&\gray{\text{Let }u(x)=\dfrac{1}{x}\text{, }v(x)=e^x} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x)&&\gray{\text{The chain rule}} \\\\ &={e^{^{\frac{1}{x}}}}\left({-\dfrac{1}{x^2}}\right) \\\\ &=-\dfrac{e^{^{\frac{1}{x}}}}{x^2} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(e^{^{\frac{1}{x}}}\right)=-\dfrac{e^{^{\frac{1}{x}}}}{x^2}$.